Senior 1 math inequalities

m<=(a+b+c)(1 a+1 b+1 c)m<=3+b a+c a+a b+c b+a c+b c because b a+a b>=2, a=b, c=2b, c=2a=2b

Therefore, m<=3+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+

The original inequality becomes (a b c) a (a b c) b (a b c) c>=m, and for this equation to hold, then m is less than or equal to the minimum value of (a b c) a (a b c) b (a b c) c, and only its minimum value is required, and the maximum value of m is it. Like this kind of problem, generally when a=b=c, take the maximum value, but this problem also has the condition a 2 b 2=c 2, then a=b takes the maximum value, then c = root number a, and the substitution can be calculated as (a b c) a (a b c) b (a b c) c is 5 3 * the maximum value of the root number. Note:

From the problem we get -b ax+1 b;

i.e. -b-1 ax b-1;

0;by -1 x 5;Gets -a ax 5a;

b-1=-a;b-1=5a;

Get a=; Not because a>0;

0;by -1 x 5;5a ax -a;;

5a=-b-1;-a=b-1;

a=；Establish;

a+b=1；

Solution: ax+1 b

b-1<=ax<=b-1

The solution set is -1 x 5

When a>0, (b-1) a=5;(-b-1) a=-1a=<0, b=<0 is not true.

When a>0, (b-1) a=-1;(-b-1) a=5a=,b= is true.

So a+b=1

-b<=ax+1<=b

b-1<=ax<=b-1

If a>0

b-1/a=-1

b-1/a=5

a=-1 2 (rounded).

If. a<0

b-1=-a

b-1=5a

a=-1/2

3) a=0 is not considered.

a+b=1

If x belongs to (-infinity, -3]:

1-x+(-x-3)>5

2x>1

x<-1 2, so it should be (-infinity, -3]:.

If x belongs to (-3,1]:

1-x+x+3>5

Look at the hypothesis.

If x belongs to (1, + infinity):

x-1+x+3>5

2x>3

x>3/2

Conclusion: x>3 2 or (-infinity, -3] is the solution.

The meaning of this problem is the set of points on the x-axis whose sum of the distances to 1,-3 is greater than 5. Since 1-(-3)=4, this point is x< or x>